How many moles and numbers of ions of each type are present in the following aqueous solutions?

a) 98.7 mL of 2.24 M aluminum chloride





b) 2.52 L of a solution containing 4.23 g/L sodium sulfate





c) 80.1 mL of a solution containing 2.96 1022 formula units of magnesium bromide per literHow many moles and numbers of ions of each type are present in the following aqueous solutions?
(a) (0.0987 L)*(2.24 M) = 0.221 mol


0.221 mol of AlCl3 contains 0.221 mol of Al(3+) ions, and 0.663 mol of Cl- ions.


Now, 0.221 mol of Al(3+) contains 0.221*6.022x10^23 ions (you do the math, and the rest of the conversion from moles to number of ions).


(b) (2.52 L)*(4.23 g/L) = 10.66 g


10.66 g of Na2SO4 (molar mass 142.04 g/mol) is (10.66/142.04) mol or 0.0750 mol Na2SO4.


0.0750 mol Na2SO4 contains 0.150 mol of Na+ ion and 0.0750 mol of SO4(2-) ion.


(c) (0.0801 L)*(2.96x10^22 /L) = 2.37x10^21


2.37x10^21 of MgBr2 contains 2.37x10^21 of Mg(2+) ion and 4.74x10^21 of Br- ion.

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